M3 June 2013 (R) Q1

EdexcelOld spec7 marksHorizontal Circular Motion

Figure 1: hollow cone with vertex downwards, base radius a, height h — Figure 1

A hollow right circular cone, of base radius \(a\) and height \(h\), is fixed with its axis vertical and vertex downwards, as shown in Figure 1. A particle moves with constant speed \(v\) in a horizontal circle of radius \(\dfrac{1}{3}a\) on the smooth inner surface of the cone.

Show that \(v = \sqrt{\left(\dfrac{1}{3}hg\right)}\). (7)

Diagram: cone with semi-vertical angle alpha, particle at radius a/3 with reaction R at angle beta to the vertical and weight mg — Mark scheme

Scheme	Marks
Vertical: \(R\cos\beta = mg\)	M1A1
Horizontal: \(R\sin\beta = \dfrac{mv^2}{r} = \dfrac{3mv^2}{a}\)	M1A1
Divide: \(\tan\beta = \dfrac{3mv^2}{amg}\)	M1dep
\(\tan\beta = \dfrac{h}{a}\)	B1
\(\dfrac{3mv^2}{amg} = \dfrac{h}{a},\ \ \dfrac{3v^2}{g} = h,\ \ v = \sqrt{\dfrac{hg}{3}}\) AG	A1
	(7)
	(7 marks)